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Abstract Cosmology And Symmetry Theory!
By!
Ian Beardsley!
Copyright © 2023"
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Table of Contents!
Abstract…………………………………………………………………..3!
Preface……………………………………………………………………4!
1.0 Six-fold Symmetry…………………………………………………..5!
2.0 Eight-fold Symmetry………………………………………………..17!
3.0 The Age of the Universe…………………………………………….22!
4.0 The Second Is The Metric…………………………………………..26!
5.0 The Proton Charge…………………………………………………..31!
6.0 Quarks and 8 and 9 Symmetry…………………………………….33!
7.0 The Solar Magnetic Field……………………………………………43!
Appendix 1 Kinetic Energies Moon and Earth…………………………55!
Appendix 2 Van Der Waals Radius……………………………………..56!
Appendix 3 Chandrasekhar Limit……………………………………….59!
Appendix 4 Cosmology Overview………………………………………62"
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Abstract!
I call it Abstract Cosmology because the approach parallels the approach taken in Abstract
Algebra, which is the way my brain works, and I see a great deal of power and elegance in this
approach. In Abstract Algebra by David S. Summit and Richard M. Foote they write “It is
important to realize, with or without the historical context, that the reason the abstract
definitions are made is because it is useful to isolate specific characteristics and consider what
structure is imposed on an object having these characteristics.”!
In section 1.0 I develop a six-fold basis for reality founded on a constant k that bridges the
microcosmos to the macrocosmos, and this is found by developing a definition for an
intermediate mass midway between protons and stars. It is necessary to develop an equation
of state for the periodic table of the elements, and we find with this approach we can predict
the radius of a proton, and discover the six-fold basis of reality, which turns out to be rooted in
biological life chemistry. In section 2.0 we touch on eight-fold symmetry and note indeed music
from around the world has been based on six-fold and eight-fold, and that it is a current
endeavor, and has been for some time, to find the underlying principles common to both, so as
to take music to the next level. I see the mathematics of reality, Nature, and the Cosmos as
music. In section 3.0 we find our theory in abstract terms suggests the lifespan of the Universe
in the Standard Friedmann Model. In section 4.0 we talk about the strange but wonderful thing
that the duration of a second, which simply came about from how we divided the rotation
period of Earth, and its orbital period, along with that of the Moon into a way of measuring
time, that the duration of a second is a Natural Metric with regard to our description of the
proton, and of the Earth-Moon-Sun orbital mechanics, and that in the end conveniently
measures the lifespan of the Universe, and the length of an Earth day and of an Earth Year.
Section 5.0 determines the charge of a proton with our constant k, and section 6.0 shows our
theory predict the description of quarks by 8 and 9 fold symmetry. Section 7.0, shows the
magnetic field of the Sun may be rooted in the six-fold symmetry of reality and may have a
function with regards to life on Earth. Laced throughout this work is the continual suggestion of
a “Genesis Project” which was a scientific fantasy of a Star Trek movie where the secrets of
life-bearing systems are discovered and allow us to convert dead worlds into living worlds.!
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Preface
One of things I wanted to go into in this paper but didn’t is that you can speak of the structure of
the solar system even though it changes with time. This is important to understand when I refer
to sizes the Moon and the planets, and their orbital distances.
The whole object of developing a theory for the way planetary systems form is that they meet the
following criterion: They predict the Titius-Bode rule for the distribution of the planets; the
distribution gives the planetary orbital periods from Newton’s Universal Law of Gravitation. The
distribution of the planets is chiefly predicted by three factors: The inward forces of gravity from
the parent star, the outward pressure gradient from the stellar production of radiation, and the
outward inertial forces as a cloud collapses into a flat disc around the central star. These forces
separate the flat disc into rings, agglomerations of material, each ring from which a different
planet forms at its central distance from the star (it has a thickness). In a theory of planetary
formation from a primordial disc, it should predict the Titius-Bode rule for the distribution of
planets. When I speak of the state of the solar system I speak of this point toward which the
solar system has formed and not the small changes that happen over millions of years due to
mutual interference between the bodies. In fact, mutual interference has torn apart possible
forming planets resulting in the current distribution we have today, because the current
distribution is more or less stable. The asteroid belt is a good example of this — it is a location
where a planet cannot form due to harmonic (repetitive) action on the orbital period at its
distance by orbital periods of planets beyond it. In short we take the state of the solar system as
an inflection point between what it became, and what it might minutely be going away from in
billions of years after it dies and can no longer support life.
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1.0 Six-fold Symmetry!
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The natural constants measure the properties of space and time. We can list some of them here:
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
"
m
P
: 1.67262 × 10
−27
kg
h : 6.62607 × 10
−34
J ⋅ s
r
p
: 0.833 × 10
−15
m
G:6.67408 × 10
−11
N
m
2
kg
2
c : 299,792, 459m /s
α : 1/137
q
p
= q
e
= 1.6022E − 19coulom bs
k
e
= 8.988E 9
Nm
2
C
2
of 7 64
We suggest the property of matter to have inertia, which is to say it resists change that the
particle a proton, a fundamental unit of matter, is a the cross-section of a 4D hypersphere in 3D
space. As such one could consider inertia as the “friction” of space given by the normal vector
holding the space-bubble in a lower dimensional “plane”:!
Thus the mass of a proton is given by the force of space measured by G, the energy given
to it given by h, the speed at which things happen c, and the surface area of this sphere
that is the cross-section of the hypersphere and the fine structure constant squared
because!
Because the fine structure constant squared is the ratio of the potential energy of an electron in
the first circular orbit to the energy given by the mass of an electron in the Bohr model times the
speed of light squared, that is it represents the ground state. It is
Equation 1.1
Thus the equation of inertia can be expressed as proton-seconds. We find that it is six-fold in
nature given the smallest integer for time (1-second) gives carbon the core element of life
chemistry, and 6-seconds is the largest integer before you get fractional protons. That is
Equation 1.2.
Equation 1.3.
m
p
4πr
2
p
α
2
α
2
=
U
e
m
e
c
2
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton ⋅ 6secon d s = hydrogen(H )
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#
From which instead of saying the left sides of these equations are seconds, we say they are
proton-seconds by not letting cancel with the bodies of these equations on the left, but
rather divide into them, which are in units of mass, giving us a number of protons. I say this is
the biological because as we shall see our equations are based on one second is 6 protons is
carbon, and 6 seconds is one proton is hydrogen, these making the hydrocarbons which are
the skeletons of biological life. We see this is a mystery of six-fold symmetry based around
biological life in the following computer program I wrote and its output:!
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by
the mass of a proton. But these masses can be considered to cancel and leave seconds. We
make a program that looks for close to whole number solutions so we can create a table of
values for problem solving.
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest
of the elements heavier than carbon are formed by fractional seconds. These are the
hydrocarbons the backbones of biological chemistry. Here is the code for the program:
m
p
1
α
2
m
p
h4π r
2
p
Gc
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#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;
do
{
printf("By what value would you like to increment?: ");
scanf("%f", &increment);
printf("How many values would you like to calculate for t in equation 1 (no more than 100?): ");
scanf("%i", &n);
}
while (n>=101);
{
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);
}}}}
We find exactly our equation predicts the second as!
Equation 1.4 !
That this equals so perfectly one second leads us to suggest the second is a Natural Unit. We
find it is, that it is in the kinetic energy of the Moon to that of the Earth times the Earth Day
which is the Earth rotation period:!
Equation 1.5. !
But that was using the average orbital velocities. We find using aphelions and perihelions (See
Appendix 1)!
Equation 1.6. "
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
KE
moon
KE
Earth
(EarthDay) = 1.2secon d s
K E
moon
K E
earth
(Ear th Day) = 1.08second s
of 10 64
To bridge the microcosmos to the macrocosmos we introduce a constant k.!
Warren Giordano wrote in his paper The Fine Structure Constant And The Gravitational
Constant: Keys To The Substance Of The Fabric Of Space, March 21, 2019:
In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic
equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where
‘ ’ is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational
constant numerically, but neglecting any units.
This follows from what Warren Giordano noticed that
Equation 1.7
We can eliminate and introduce the 6 of six-fold symmetry by introducing Avogadro’s
Number
We make an equation of state for the periodic table:
We can say for any element
Where is the number of protons in the element, so for carbon
Thus by equation 1.7 we have
Equation 1.8 .
h
1 + α
α
10
23
h(1 + α) ⋅ 10
23
= G
10
23
N
A
= 6E 23
ℍ = 1
gr a m
atom
N
A
⋅ 𝔼 = 6E 23
𝔼
N
A
=
Z ⋅ 6E 23proton s
Z ⋅ gr a m s
𝔼 =
Z ⋅ gr a m s
Z ⋅ proton s
Z
ℂ
ℂ =
6gr a m s
6proton s
N
A
=
6(6E 23proton s)
6gr a m s
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
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In the next step to bridge the microcosmos to the macrocosmos we create an intermediary
mass midway between the proton mass and the upper limit of mass for a white dwarf
star to form without collapsing into a blackhole star. It balances with its gravity by radiation
pressure alone, the so-called Chandrasekhar limit (See Appendix 3 for how it is derived):!
Equation 1.9
We make the approximation and define with the geometric mean:
Equation 1.10
Giving
Equation 1.11
Thus by equation 1.8 and equation 1.11 and using , we have
Equation 1.12
We had
We can hone this by reintroducing 0.77 for 3/4
Thus precisely:
We have
We have honing our
m
i
m
p
M ≤ 0.77
c
3
ℏ
3
G
3
N
m
4
p
= 1.41 ⊙
0.77 ≈ 3/4
m
i
m
i
= Mm
p
m
i
=
3
2
c
3
ℏ
3
G
3
m
2
p
1/2
ℏ = h /2π
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
m
i
=
3
2
c
3
h
3
8π
3
G
3
m
2
p
1/2
= 67.9943kg
3
2
= 0.8660
0.77 = 0.8775
0.8775
0.8660
= 1.01328
m
i
m
i
= (67.9943)(1.01328) = 68.897kg ≈ 69kg ≈ 70kg
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Thus since we said with our estimate
We have a honed value of
This is our constant for bridging the microcosmos with the macrocosmos. We define the Earth
to be the ground state for the solar system yielding our six-fold basis for Nature
Equation 1.13.
Where is the average orbital velocity of the Earth.
We note the ground-state for an electron in a hydrogen atom gives
Where is the potential energy of the electron and is its rest energy.
Thus we have the radius of a proton is given by carbon by evaluating at one second:
But to get that we have to multiply by one second and we need one second in terms of the atom
for a theory of the proton. I find we can do that…
Substitute for to get
k =
1
m
2
i
h
(1 + α)
G
⋅ N
A
𝔼 =
1
773.5
⋅
s
m
k =
1
(68.897kg)
2
(6.62607E − 34) ⋅
(1.007299)
6.67408E − 11
⋅ 6.02E 23 = 0.001268291s /m
1
k
= 788.4626
m
s
k v
e
= 6
k v
e
=
29790m /s
788.46m /s
= 6.145748
v
e
α
2
=
U
e
m
e
c
2
U
e
m
e
c
2
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
t
6
=
1
α
2
m
p
h 4π r
2
p
Gc
t
6
=
r
p
α
2
m
p
h 4π
Gc
R
H
/2
r
p
of 13 64
Where is the Van Der Waals radius for a hydrogen atom (See Appendix 2). We have
now introduced the radius of a hydrogen atom . Our formulation of inertia as
proton seconds is a form of impulse. To change that to momentum we have to divide by a
second. This radius of the hydrogen atom is the Van Der Waals radius, which is the closest
distance between two hydrogen atoms noncovalently bound. It is 120 pm. Divide that by
where 1/k is our constant
And we find
We have our equation for the radius of a proton
We only need to multiply it by to have the right units, and we get
Equation 1.14.
Then suggest we picked up 9/8 in approximations which is close to one anyway so we write
Equation 1.15.
We form constants:
t =
R
H
2α
2
m
p
h 4π
Gc
R
H
R
H
= 1.2E − 10m
ck
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
1 + α
⋅
1
N
A
𝔼
≈
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
N
A
𝔼
t
ck
=
3 2
16
⋅
6.626E − 34
6.674E − 11
⋅
18769
(1.6726E − 27)
2
⋅
1.2E − 10
6.02E 23
⋅
1
π
= 1.12secon d s
3 2
16
⋅
h
Gπ
⋅
1
α
2
m
2
p
⋅
R
H
N
A
𝔼
= 1.12secon d s
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
t /ck = 1secon d
r
p
=
9
8
2 ⋅
1
m
p
hc
4π
3
G
⋅
R
H
N
A
𝔼
r
p
=
1
m
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
r
p
= 8.26935E − 16m ≈ 0.827f m
r
p
=
1
kg
(
kg
m
2
s
)
(
m
s
)
(
kg
s
2
m
3
)
R
H
N
A
ℍ
=
1
kg
(
kg
)
m
of 14 64
And we have the Equation:
Equation 1.16
We can say that Avogadro’s number is not an arbitrary number because it is such that there are
twelve grams of carbon and carbon is 6 neutrons plus 6 protons equals 12. We now want to add
to this section the equation of the mass of a proton.
Equation 1.17
If our equation is right and we put it into natural units then the product should be close to
one:
Let us start with the units with which we are working:
And convert these to proton-masses and proton-radii:
Now we find k in these units:
k =
hc
2π
3
G
= 6.93E − 9kg
R
H
N
A
𝔼
= 1.99E − 34m
r
p
m
p
= k
R
H
N
A
𝔼
m
p
=
1
r
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
r
p
m
p
G =
m
3
kg ⋅ s
2
h = kg ⋅
m
2
s
c = m /s
G = 6.67408E − 11
m
3
kg ⋅ s
2
⋅ 1.67262E − 27kg(0.833E − 15m)
3
⋅ s = 193,131, 756
h = 6.62607E − 34kg ⋅
m
2
s
⋅
s
(0.833E − 15)
2
(1.67262E − 27kg)
= 5.71E 23
c =
(299,792, 459m /s)(1sec)
(0.833E − 15m)
= 3.6E 23
R
H
=
1.2E − 10m
0.833E − 15m
= 144,058
of 15 64
Thus we have:
Equation 1.18.
!
We will say our theory is holistic and that we are describing the proton radius in terms of the
whole of which it is a part, namely, the radius of a hydrogen atom, more specifically the Van Der
Waals radius, which is determined by hydrogen gas, or (See Appendix 3 for the theory of the
Van Der Waals radius).
We have that
Equation 1.19
Equation 1.20
Which give the core life elements carbon and hydrogen, which are also the hydrocarbons that
make-up the skeletons of biological life chemistry:
Equation 1.21
Saying one second gives 6 protons is carbon (C) and, six seconds gives 1 proton is hydrogen (H):
Equation 1.22
We have the six-fold symmetry in the constant k and the orbital velocity of Earth:
Equation 1.23
We substitute Equation 1.20 and 1.23 into equation 1.21 giving:
k =
hc
2π
3
G
= 6.93E − 9kg
k =
(5.71E 23)(3.6E 23)
2π
3
(193131756)
= 4E18proton − m a sses
r
p
m
p
= k
R
H
N
A
𝔼
r
p
m
p
=
(4E18)(144058)
(6E 23)
=
5.76E 23
(6E 23)
= 0.96 ≈ 1
H
2
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 1secon d
K E
Earth
K E
moon
(Ear th Day) = (1secon d )
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton ⋅ 6secon d s = hydr ogen(H )
k v
e
= 6
of 16 64
Equation 1.24
Or we can let cancel with the numerator and not write in protons:
Equation 1.25
We see the radius of a proton to the mass of a proton is proportional to the kinetic energy of the
Moon to the kinetic energy of the Earth, times the Earth rotation period (EarthDay) times the
square root of the Earth’s orbital velocity times k. This may turn out to be a core equation of a
so-called “Genesis Project” where we try to change dead worlds into living worlds, or maintain
living worlds as was fantasized in Star Trek, the movie. But the orbital velocity of the Earth is
given by the Sun’s mass and Earth’s distance from the Sun:
Equation 1.26
But the Moon perfectly eclipses the Sun which means as seen from the earth the Moon appears
to be the same size as the Sun. This is because
Equation 1.27
Where is the Earth orbital distance, is the Moon’s orbital distance, is the Sun’s radius,
and is the Moon’s radius.. Thus in terms of the moon and the Sun
Equation 1.28
So we have from 1.26
Equation 1.29.
This means from 1.25
Equation 1.30
1
α
2
⋅
r
p
m
p
h 4π
Gc
⋅
K E
earth
K E
moon
⋅
1
Ear th Day
= k v
e
⋅ proton s
m
p
1
α
2
⋅
r
p
m
p
h 4π
Gc
=
K E
moon
K E
earth
(Ear th Day) k v
e
v
e
=
GM
⊙
r
e
r
e
r
m
≈
R
⊙
R
m
r
e
r
m
R
⊙
R
m
r
e
=
R
⊙
R
m
r
m
v
e
= G
M
⊙
R
⊙
⋅
R
m
r
m
1
α
2
⋅
r
p
m
p
h 4π
Gc
=
K E
moon
K E
earth
(Ear th Day) k G
M
⊙
R
⊙
⋅
R
m
r
m
1/2
of 17 64
2.0 Eight-fold Symmetry
of 18 64
Since in equation 1.13 of the previous section and is dimensionless, we can write in
the 6 of sixfold symmetry for it:
Equation 2.1.
We see that
Where hydrogen and carbon are the central structural elements of life composing the
hydrocarbons which are the skeletons of biological chemistry.
Thus
Is a quantity that maps things to things. We find since the orbital velocity of the moon
, we have
We have equation 1.11
Which honed is
k v
e
= 6
1
α
2
⋅
r
p
m
p
h 4π
Gc
=
K E
moon
K E
earth
(Ear th Day) k v
e
1
α
2
⋅
r
p
m
p
h 4π
Gc
= 6
K E
moon
K E
earth
(Ear th Day)
1
α
2
4πh
Gc
:
r
p
m
p
↦ 6secon d s
1
t
1
⋅
1
α
2
⋅
r
p
m
p
h 4π
Gc
= 6proton s = Carbon
1
t
6
⋅
1
α
2
⋅
r
p
m
p
h 4π
Gc
= 1proton = Hydr ogen
(
r
p
m
p
)(
1
α
2
4πh
Gc
)
=
(
4.98E11
m
kg
)(
1.21E − 11kg
s
m
)
= 6secon d s
1
α
2
4πh
Gc
= 1.21E − 11kg
m
s
v
m
= 1,022m /s
(1.21E − 11kg ⋅
s
m
)(1022m /s) = 1.23662E − 8kg
m
i
=
3
2
c
3
ℏ
3
G
3
m
2
p
1/2
of 19 64
which gives gives
Thus we have
Six-fold which is carbon (C) has map the radius to mass of a proton to the kinetic
energy of the Moon to the kinetic energy of the Earth by a factor of the Earth period of rotation,
which is 1 second and that the intermediate mass , is 8 times the factor that does that
mapping.
Where 1 second is
Giving the radius of a proton as
From
Where is six seconds. What is the comparison of six-fold symmetry to eight-fold symmetry?
Six-fold is in the 18 groups of the periodic table that organizes elements according to their
properties. Six-fold is dynamic because 2 and 3 are the smallest prime numbers so they are the
smallest factors down to which any number can be reduced. We have
, , ,
m
i
= (67.9943)(1.01328) = 68.897kg ≈ 69kg ≈ 70kg
68.897kg
1.23662E − 8kg
= 8.0865585 ≈ 8
m
i
(
1
α
2
4π
Gc
v
m
)
= 8
1
6
⋅
r
p
m
p
1
α
2
h 4π
Gc
=
K E
moon
K E
earth
(Ear th Day) = 1secon d
1
α
2
4πh
Gc
r
p
m
p
m
i
m
i
= 8
(
1
α
2
4π
Gc
v
m
)
3 2
16
⋅
h
Gπ
⋅
1
α
2
m
2
p
⋅
R
H
N
A
𝔼
= 1secon d
r
p
=
1
m
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
1
t
6
⋅
1
α
2
⋅
r
p
m
p
h 4π
Gc
= 1proton
t
6
2 ⋅ 3 = 6
3
2
= 9
2 ⋅ 9 = 18
3 ⋅ 6 = 18
of 20 64
That is dynamic. That is how you play music in 6/8 that resolves in . It is in the
Flamenco of Spain to play 3 groups of six to make 18 like in the periodic table of the elements,
then strum another group of chords six beats long usually with rhythm 2+2+2, to make 24 in
order to come around to cycles of 12. This form is called Bulerias, and is considered the most
dynamic in the genre. However, just as popular is eight, the tango flamencos, and the rumba,
which is 4+4+4+4=16. Other forms are fandangos (four groups of 3 making 12) and Sevillanas,
the same thing. Middle Eastern music has much along these lines and so does Hindu Indian
music which are North Indian Classical music (Mostly tin tal which is four groups of four to
make 16) and Ghuzals, romantic music played in 6.
12 = 2 ⋅ 6
of 21 64
of 22 64
3.0 The Age of the Universe
of 23 64
We have the possible core equation for a so-called Project Genesis giving the ratio of the radius
of a proton to its mass the kinetic Energy of the moon to that of the Earth:
We now see the same equations predict the age duration of the Universe in the standard model
(See Appendix 4). That is we start with the following two equations:
Eq 3.1.
Eq. 3.2
We want to factor 3.1 into a ratio between kinetic energies times a time as we have in 3.2. We
write 3.1 as
And noticing
Which gives
Is
And we know
And have
1
α
2
⋅
r
p
m
p
h 4π
Gc
= ⋅
K E
moon
K E
earth
(Ear th Day) k v
e
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 1secon d
K E
moon
K E
earth
(Ear th Day) = 1secon d
6α
2
r
p
m
4
p
⋅ Gc
4πh
=
m
p
1secon d
m
3
p
Gc
h
= KE
36α
4
⋅
m
p
⋅ K E
4π
=
r
2
p
m
2
p
1secon d
K E =
4π
36α
4
⋅
r
2
p
m
p
t
2
1
t
1
=
1
6α
2
m
p
h 4π r
2
p
Gc
4π
36α
2
⋅
(0.833E − 15)
2
t
2
1
⋅ (1.67262E − 27) = 1.413E − 49J
of 24 64
We also notice that obviously by dimensional analysis
We have the kinetic energy and we need another kinetic energy to form the ratio:
We have the intermediary mass and the constant k inverted. We can write
This is
=3.03E56 seconds
But we also formulated
So we can write
m
3
p
Gc
h
=
4π
36α
4
⋅
r
2
p
m
p
t
2
1
h
G
⋅
c
3
m
p
= KE
h
G
⋅
c
3
m
p
(K E
moon
)
= 4.6654
K E = m
3
p
Gc
h
m
3
p
⋅
Gc
h
K E
(t im e) = aT im e
m
i
K E = m
i
(
1
k
)
2
K E = (68.897kg)(788.4626m /s)
2
= 42,831,358J oules
K E
1
K E
2
(t im e) = (aT ime)
m
i
(
1
k
)
2
m
3
p
Gc
h
(1secon d ) =
(42,831,358J )
(1.4130E − 49J )
(1secon d )
6.62607E − 34
6.67408E − 11
⋅
299792459
3
1.67262E − 27
= 1.599298E 29J
h
G
⋅
c
3
m
p
= 1.599298E 29J
of 25 64
1 Earth year = (365.25)(24)(60)(60)=31557600 seconds
The universe is theorized by standard models to die in 100 trillion years, which is when the last
stars born will die out. This is exactly 1E14 years
We see the moon is connected to how old the universe is theorized to become.
h
G
⋅
c
3
m
p
m
i
(
1
k
)
2
(1secon d ) =
1.599298E 29J
42831358J
= 3.734E 21secon d s
3.734E 21s
31557600s
= 1.1832332E14years ≈ 1E14years
of 26 64
4.0 The Second Is The Metric
of 27 64
We have !
4.1. !
4.2.
, and . In both 4.1 and 4.2 we have
kinetic energy over kinetic energy times one second equal to a pivotal time period in Nature, the
first the Life Span of the Universe, and the second the Earth Day, period of rotation of the Earth.
In equation 4.1 is given by the white dwarf star:
And
Is the Chandrasekhar limit for the mass of a star to not collapse into a blackhole so it can
become a white dwarf star. We can write 4.1
4.3.
Using where .
We have
EarthDay=(24)(60)(60)=86400 seconds
LifeSpanUniverse=1E14 Years = 3.734E21 seconds
KE Moon=3.428E28J
KE Earth=2.7396E33J
Kinetic energies used from aphelions and perihelions (See Appendix 1). We have
=
h
G
⋅
c
3
m
p
m
i
(
1
k
)
2
(1second ) = Li feSpa nUniverse
K E
Earth
K E
moon
(1secon d ) = Ear th Da y
h
G
⋅
c
3
m
p
= Kinet icE nerg y
m
i
(
1
k
)
2
= Kinet icE nerg y
m
i
m
i
= Mm
p
M = 0.77
c
3
h
3
8π
3
G
3
m
4
p
k
4
m
3
p
⋅
h
2
c
6
G
2
(
Ear th Day
Li feSpa nUniverse
)
2
(
K E
moon
K E
earth
)
2
= M
WhiteDwar f
m
i
= Mm
p
M = M
WhiteDwar f
k
4
m
3
p
⋅
h
2
c
6
G
2
=
(
1
788.4626
)
4
1
(1.6726E − 27)
3
(6.626E − 34)
(6.674E − 11)
2
(299792459)
6
(2.58747E − 12)(2.1371E 80)(9.926581E − 47)(7.25979E50) = 3.96631k ilogram s
of 28 64
Using our honed value for k.
=(5.354E-34)(1.5657E-10)=8.3827578E-44
Multiplying these last two:
(3.96631kg)(8.3827578E-44)=3.32486E30 kg
This mass should equal that of a white dwarf star (Its lower limit not to collapse into a
blackhole). There can be some play as we are suggesting the moon is an indicator somehow of all
that is going on in the Universe, and it only has to be at approximately the right orbit to do this.
We have
Which makes the equation pretty accurate in this form considering we are comparing something
like the Earth day to something as immense as the LifeSpanUniverse. We are going to want to
look at:
Where is the Chandrasekhar limit.
4.4.
Thus we have
Where is a white dwarf star. We have
We suggest for some mass , we have
(
Ear th Day
Li feSpa nUniverse
)
2
(
K E
moon
K E
earth
)
=
(
86400
3.734E 21
)
2
(
3.428E 28J
2.7396E 33J
)
2
2.86416E 30kg
3.32486E 30kg
= 86 %
h
G
⋅
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
M
c
h
G
⋅
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
h
G
⋅
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
M
c
K E
Earth
K E
moon
(1secon d ) = Ear th Da y
M
of 29 64
It is given by
4.5
For all practical purposes this is the mass of the Moon, which is exactly 7.34767E22kg.
Conclusion next page…
h
G
⋅
c
3
m
p
M
(
1
k
)
2
(1secon d ) = 1Ear thYear
M =
h
G
⋅
c
3
m
p
(1Ear thYear )
(
1
k
)
2
(1secon d )
h
G
⋅
c
3
m
p
=
6.62607E − 27
6.67408E − 11
⋅
(299792459)
3
1.67262E − 27
= 1.599E 36J ≈ 1.6E 36J
1
k
2
= (788.4626)
2
= 621,673.27
1Ear thYear = (365.25)(24)(60)(60) = 31557600s
M =
1.599E 36
621673
⋅
1
31557600
= 8.15E 22kg
8.15E 22kg
7/34767E 22kg
= 1.10919516 ≈ 1.12m oon s
of 30 64
Thus we have
Where is a white dwarf star and gives LifeSpanUniverse.
Where Earth-Moon-Sun orbital parameters give the Earth rotation period.
Where is the mass of the moon gives the Earth orbital period. We have already formulated
From the radius of a proton and the mass of proton , our theory panned out: The second is
a unit of time that expresses the cycles of the Universe, it is the metric (measures things
conveniently).
h
G
⋅
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
M
c
K E
Earth
K E
moon
(1secon d ) = Ear th Da y
h
G
⋅
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
M
m
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 1secon d
r
p
m
p
of 31 64
5.0 The Proton Charge
I construct the electric field as such (Fig. 2): is one component of . Events in
are through time with components and where c is the speed of light. The
electrons and the protons and in the presence of one another cross into giving
themselves acceleration in the and directions.
We have
We suggest
Equation 5.1
The position of changes in the space of and has travelled to a place in
time, as well. Since ct=meters there is a g such that . Thus since
x
ℝ
3
(x, y, z)
ℝ
3
(x, y, z)
ct
1
ct
2
q
e
q
p
ct
1
ct
2
x
−x
i
j
k
0 ct
1
0
0 0 ct
2
= (ct
1
− ct
2
)
i
··
x ∝ (ct
1
− ct
2
)
i
x
ℝ
3
(x, y, z)
t (0,ct
1
, ct
2
)
··
x = g(ct
1
− ct
2
)
i
g ⋅ m =
m
s
2
Fig. 2
of 32 64
and that is . Thus g is frequency squared ( ) and we suggest it is
derived from the separation between the charges and . That there is some velocity v such
that . We call upon our equation for 1 second:
Letting v= from our equation for k
And we have
Equation 5.2
We get
Equation 5.3 "
g =
1
s
2
=
1
t
2
1
s
2
⋅
m
s
⋅ s =
m
s
2
f
2
q
e
q
p
f =
(
v
x
)
r
p
m
p
h 4π
Gc
= 1secon d
(
v
x
)
2
(c)
r
p
m
p
h 4π
Gc
=
k
e
m
⋅
q
2
x
2
v
2
(c)
r
p
m
p
h 4π
Gc
= k
e
q
2
m
α
2
6
⋅
1
k
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼 =
1
773.5
s
m
1
k
= 773.5
m
s
m = m
p
c
k
e
(
α
2
6
⋅
1
k
)
2
h 4π r
2
p
Gc
= q
2
h 4π r
2
p
Gc
=
(6.626E − 34)(4π)(0.833E − 15)
2
(6.674E − 11)(299,792, 459)
= 5.37E − 31
q = (0.033)
1
36(18769
2
)
773.5
2
(5.37E − 31) = 9.1435E − 19C
q
q
p
= ⋅
9.1435E − 19C
1.602E − 19C
= 5.71proton s ≈ 6protons
of 33 64
6.0 Quarks and 8 and 9 Symmetry!
of 34 64
We have seen our constant k, and intermediary mass described the periodic table of the
elements in our equation of state for it!
Which is
But that they predict the radius of the proton which makes up these elements, and its charge
!
We further find the constant k bridges the microcosmos (atoms and protons) with the
macrocosmos, the Earth-Moon-Sun System in in terms of sixfold symmetry as well!
!
m
i
N
A
=
Z ⋅ 6E 23proton s
Z ⋅ gr a m s
𝔼 =
Z ⋅ gr a m s
Z ⋅ proton s
N
A
⋅ 𝔼 = 6E 23
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
m
i
=
3
2
c
3
h
3
8π
3
G
3
m
2
p
1/2
= 67.9943kg
k =
1
m
2
i
h
(1 + α)
G
⋅ N
A
𝔼 =
1
773.5
⋅
s
m
r
p
=
1
m
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
c
k
e
(
α
2
6
⋅
1
k
)
2
h 4π r
2
p
Gc
= q
2
q
q
p
= ⋅
9.1435E − 19C
1.602E − 19C
= 5.71proton s ≈ 6protons
k v
e
= 6
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
K E
moon
K E
earth
(Ear th Day) = 1.08secon d s
of 35 64
The same sixfold symmetry we see in the hydrocarbons that are the structural skeletons of life
chemistry
And ultimately that the Moon describes the cycles of the universe in terms of all of this:
And we see the duration of a second is a Natural Metric. But now we see that k and determine
the structure found in the subatomic particles, the quarks, and that this predicts their associated
eightfold and ninefold symmetries.!
The eightfold is a basic and dynamic structure of Nature used both in describing subatomic
particles as in the eightfold way and chemical compounds as in the octet rule. See diagrams in
the following pages. The hadron class of subatomic particles is described by eightfold
symmetry where kaons, pions, and eta mesons are on a regular hexagon (sixfold) to describe
an eightfold categorization for mesons. And the the same same for sigma baryons, the lambda
baryon, and the xi baryons in the baryon octet.!
An atom has core electrons and valence electrons. Valence electrons used for bonding, are the
outermost electrons in the highest s and p subshells. The octet rule states that atoms bond
such that each atoms acquires eight electrons in the outermost shell. This allows us to make
electron diagrams Hydrocarbons are the skeletons of life chemistry. The simplest hydrocarbon
is methane CH4. We see the eight outermost electrons of carbon which is the core atom form 4
pairs, each providing two electrons for hydrogen which holds two electrons in its outermost
orbit. We said that:!
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton ⋅ 6secon d s = hydr ogen(H )
h
G
⋅
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
K E
Earth
K E
moon
(1secon d ) = Ear th Da y
h
G
⋅
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 1secon d
m
i
of 36 64
Is sixfold symmetry in the hydrocarbons. We now have eightfold symmetry!
That
There is a strong connection between sixfold and eightfold. Angles A in a polygon
We have 180 degrees is the angles in a triangle. For eightfold (n-2)=6 is sixfold. The 180 in a
triangle is the object into which any polygon can be broken-up into. We see intermediate mass
divided up by 8 gives the factor:
That maps into six second is the one proton is hydrogen (H) corresponds to one second is
carbon (C) these making the simplest hydrocarbon methane (CH4) of eightfold symmetry given
by compounds completing an octet. The intermediate mass divided by 8 is:
We have said the the Earth orbital velocity gives sixfold symmetry
We can determine from another value for k:
=
1
t
1
⋅
1
α
2
⋅
r
p
m
p
h 4π
Gc
= 6proton s = Carbon
1
t
6
⋅
1
α
2
⋅
r
p
m
p
h 4π
Gc
= 1proton = Hydr ogen
1
α
2
4πh
Gc
:
r
p
m
p
↦ 6secon d s
m
i
= 8
(
1
α
2
4π
Gc
v
m
)
A = (n − 2)(180
∘
)
m
i
m
i
8
=
(
1
α
2
4π
Gc
v
m
)
r
p
m
p
m
i
m
i
8
=
68.897
8
= 8.612125kg
k v
e
= 6
m
i
/8
k =
1
m
2
i
h
(1 + α)
G
⋅ N
A
𝔼
1
(8.612)
2
⋅ (6.62607E − 34) ⋅
1.007299
6.67408E − 11
⋅ 6.02E 23 = 0.081172975 ≈ 0.081173
of 37 64
We have that the Moon and Earth together describe physical systems (even on the atomic level):
Where is the orbital velocity of the Moon and we see then the moon produces ninefold
symmetry where the earth produced sixfold.
As it would turn out ninefold symmetry was used to for quark theory. Ninefold symmetry is very
compatible with sixfold symmetry because
and
This is the most dynamic of relationships in music around the world for playing in six and is
probably why the periodic table of the elements is cyclical over 18 groups in describing the
properties of the elements. Ultimately the eightfold way gave rise to the ninefold symmetry for
the modern theory of quarks, the subatomic particles from which atomic particles are made.
They call their description as given by nonets.
By 1947 the smallest particles were electrons, protons, neutrons, and photons that make up
everyday experience, but there were some exotic particles in cosmic rays, rays that come to us
from space, the pions and muons and the neutrino was hypothesized. As well there was the
suggestion of antimatter with the discovery of the positron. The eightfold way organized these
particles but it inspired a quark model that suggested mesons should be grouped in 9, where
mesons are a type of hadron, which are particles composes of an equal number of quarks and
antiquarks, and hadrons are subatomic particles held together by the strong interaction in
contrast with electric fields for the interactions of protons and electrons.
The eightfold way was given to us by Murray Gell-Mann and Yuval Ne’eman in 1961. It organizes
particles in octets either the mesons or baryons and this by the properties of spin and a new
property that is conserved called “strangeness” given to us by Gell-Mann, Tadau Nakano, and
Kazuhiko Nishijima. The ninefold nonets for the mesons arises because if we restrict ourselves
to “light quarks”, quarks an order of magnitude less massive than other quarks, there are three:
up, down, and strange as they are called. Since mesons are composed of a quark and an
antiquark there are three ways of picking each so there are nine such mesons.
Continue to next page for illustrations…
1
k
= 12.4m /s
k v
m
= 0.081173)(1022m /s = 9.108 ≈ 9
v
m
2 ⋅ 9 = 18
18 = 6 ⋅ 3
of 38 64
of 39 64
of 40 64
of 41 64
of 42 64
We see the moon describes things on an atomic level and of macro scales like the age of the
Universe. In the end we need to consider why the Moon is such that it does this, we know it
makes life on Earth possible because it holds the Earth at its inclination to its orbit (the ecliptic)
allowing for the seasons and keeping temperature from extreme heat and extreme cold. We will
have to consider fivefold symmetry, because while sixfold describes for the most part the
physical, fivefold describes life. It contains the golden ratio."
of 43 64
7.0 The Solar Magnetic Field
We model the formation of the solar system from a slowly rotating gas cloud, a nebula of
gaseous molecules, that collapses into a flat disc with a protostar at its center. The star turns on
and blows lighter elements far away, like hydrogen and helium, from which form the gas giants,
like Jupiter and Saturn, and the heavier elements stay closer in, like iron and silicates, from
which form the terrestrial planets like Venus, Earth, and Mars form. There are basically three
factors that determine its structure, the inward gravity, the pressure gradient outward which
balances with the inward gravity, and the outward inertial forces from the planets’ orbits. The
flattened rotating disc is broken up into rings each that has a mass spread out over it from which
the planets form. We estimate the ring associated with the Earth, had in its lower limit 230 earth
masses spread over it for the Earth to form. We further estimate that the Venus ring had a mass
spread over it of 230 Venus masses for Venus to form, and the Mars ring similarly had 230 Mars
masses spread out over it for Mars to form. The asteroid belt had about 200 of it masses, and the
Jovian planets 5, 8, 15, and 20 masses of each respectively. For Mercury it requires a factor of
about 350 because it is mostly iron condensations with incomplete silicon condensations.
Plotting these logarithmically we get the exponent of r, the distance of a planet from the sun is
-1.5 so that the density distribution of the protoplanetary disc is:
Giving a mass
With pressure gradients playing the key role in the formation of solar system, less attention is
payed to the magnetic field of the Sun. However, in the older literature, one of the pioneer’s of
this aspect found something very interesting concerning it. He was Alfven (1942). At the time
people were suggesting instead of the solar system forming from a rotating nebula, rather the
sun came into existence not at the same time at the center of the disc, but rather passed through
clouds and captured material after already existing. He figured for the captured material its
inward component v, and density , at a distance r from the sun, had to conserve mass, which
required:
He figured as the velocities of the atoms got closer to the sun, were moving then faster, collisions
would increase, and so temperature would go up, ionizing the atoms and therefore ionized, the
magnetic field becomes important. He considered for simplicity the solar magnetic field was
generated by a dipole moment , a vector quantity, and that a particle moving in the plane of
that vector with mass m and charge q, would have all of both the gravitational and magnetic
forces in that plane, so the problem becomes two-dimensional and required only the and , of
polar coordinates. The differential equations of its motion would be:
σ (r) = σ
0
r
−3/2
σ
0
= 3300
M =
∫
2π
0
∫
r
h
r
s
σ (r)r drdθ
ρ
dM
dt
= 4π r
2
ρv
μ
θ
r
of 44 64
Equation 7.1
And,
Equation 7.2
We can integrate equation 7.2 with the boundary condition that the angular momentum of the
particle is zero at large distances from the sun to get:
Equation 7.3
And substitute it into equation equation 7.1 for to get
Equation 7.4
Which we can write
Equation 7.5
We then integrate this with respect to r with the boundary condition that at large r and get
Equation 7.6
He then notices there is another value for which . It is
Equation 7.7
This is interesting because it means the particle can never approach the Sun closer than this
value, and it depends only on the value q/m, the charge to mass ratio of a particle. He took this
as hydrogen because ionized it is a proton, for which q/m is well defined. He estimated what the
magnetic field of the Sun could have been in this earlier stage of its life, and adjusted for the fact
that hydrogen doesn’t ionize until it reaches a velocity of 5E4 m/s and found that was the
region occupied by the major planets which are Jupiter and Saturn mostly made of hydrogen
and helium.
Certainly today we don’t see the planets as having formed from material gathered by the Sun in
its journey, but rather think the Sun and planets formed at the same time from a cloud that
collapsed into a rotating flat disc. And indeed, there may be stars in the galaxy that pass through
clouds and gather material, and indeed Alfven’s equations would hold preventing ionizing
clouds from falling into their star. But we can also apply his equation to our Sun today, for which
we know a great deal about its magnetic field, which also happens to be an important thing to
study and for which we have satellites in the Lagrange points, where the Earth’s gravity cancels
m
··
r = −
GM
⊙
m
r
2
+
qμ
·
θ
r
2
+ mr
·
θ
2
m
r
d(r
2
·
θ )
dt
= −
qμ
·
r
r
3
mr
2
·
θ = −
∫
qμ dr
r
2
=
qμ
r
·
θ
m
··
r = −
GM
⊙
r
2
+
2q
2
μ
2
m
2
r
5
·
r
d
·
r
dr
= −
GM
⊙
r
2
+
2q
2
μ
2
m
2
r
5
·
r = 0
·
r
2
=
2GM
⊙
r
−
q
2
μ
2
m
2
r
4
·
r = 0
r
c
=
(
q
2
μ
2
2GM
⊙
m
2
)
1/3
r
c
of 45 64
with that of the Sun, where the orbits are very stable, so we can understand the solar magnetic
field. It is a complex field, that interacts with the Earth’s magnetosphere, and we need to predict
solar maximums, so we have warning as to whether there will be a magnetic storm that will
knock out our electrical grid and internet, ahead of time.
During solar minimum the solar magnetic field has closed lines, that flow out one pole and into
the other. The dipole field of the sun is about 50 Gauss. There are 10,000 Gauss in a Tesla, so
that is 5E-3 Tesla. That is the magnetic field strength where the field goes into the poles. The
total magnetic field of the Sun at the Earth, is all the components taken together, which are
, , and . The important component is , because it runs north-south, so it is
perpendicular to the ecliptic, the path traced out by the sun due to the earth’s orbit. It is the
component that interacts with the Earth magnetosphere, and when it points southward, it will
connect with the Earth’s magnetosphere which points northward so the solar poles flow into the
Earth poles and the Earth field then gets disrupted allowing particles from the solar wind to rain
down along Earth magnetic field lines causing the Aurora. The solar magnetic field doesn’t
always stay around the Sun itself, but the solar wind carries it through the solar system until it
collides with the interstellar medium reaching the heliopause. Thus the Sun creates the
Interplanetary Magnetic Field (IMF) which has a spiral shape because the Sun rotates once
about every 25 days. But the upshot is that at Earth we have
Moderate Magnetic Field: 10 nT
Strong Magnetic Field: 20 nT
Very Strong Magnetic Field: 30 nT
For our purposes we want to return to equation:
B
t
B
x
B
y
B
z
B
z
of 46 64
And ask just what is , because in the time that Alfven was working we worked with magnetic
fields differently, aside from his equation uses a trick, which we still use today, and that is to
consider the magnetic field a dipole. To consider it like this is to say there are two monopoles
opposite in polarity. According to Maxwell’s equation we cannot have magnetic monopoles,
though they are predicted by some modern theories, they have never been found. The trick is in
that by treating the North magnetic pole and South magnetic pole as separate magnetic charges
is to treat them like we do electric charges, the charge of a proton and the charge of an electron,
which can be convenient for making computations, but don’t exist that we know of. So we will
solve equation 56 for , and see what its units are so we can understand what it represents and
we will let m be the mass of a proton and q the charge of a proton. We get:
Equation 7.8
We can write these units as, by taking Coulombs (C) equal to
Equation 7.9
This is units of force per current density, which makes sense because a flowing current creates a
force. We can also write it:
Equation7.10
Which is energy per magnetic field strength in that the SI units of magnetic field strength is
amps per meter. This tells us:
Equation 7.11
Thus we will use the energy as ionization of hydrogen, the energy to remove its electron and
make it a proton:
We have
, , ,
r
c
=
(
q
2
μ
2
2GM
⊙
m
2
)
1/3
μ
μ
r
3
GM
⊙
m
2
p
q
2
p
= μ =
m
3
⋅ kg
C ⋅ s
a mp ⋅ secon d s
kg
m
s
2
⋅
m
2
a mps
kg
m
2
s
2
⋅
m
a mps
μ =
En erg y
Magnet icFiel d St rength
H ⟶ H
+
+ e
−
= 1proton = 2.18E − 18J
q
p
= 1.6E − 19C
m
p
= 1.67E − 27kg
G = 6.67408E − 11N
m
2
kg
2
M
⊙
= 1.989E 30kg
of 47 64
We want to look at the Sun as having a current flowing around its equator in a loop with its
radius
We find for the dipole field of the Sun at 50 Gauss=5E-3T, which is about 100 times stronger
than the Earth magnetic field, that this is a current I=5.5362E12 amperes driving the solar
magnetic dipole. This gives us that since the Earth orbit (1AU=1.495979E11m):
Equation 7.12
Equation 7.13
Or,…
Equation 7.14
The radius of a proton is 0.833E-15m. We have that r is:
Equation 7.15
R
⊙
= 6.957E8m
I
1AU
=
5.5362E12a mps
1.496E11m
= 37.0a m ps /m
μ =
Ioni zat ion En erg y
Magnet icFiel d St rength
=
2.18E − 18J
37.0A m ps /m
= 5.892E − 20
J ⋅ m
A
r =
(
q
2
p
μ
2
2GM
⊙
m
2
p
)
1/3
= 4.9324E − 15m
r
r
p
= 5.92 ≈ 6Pr oton Ra dii = carbon
of 48 64
Our six-fold symmetry unfolding. This is again the carbon the core element of life. We see it
provided for by the Sun’s magnetic field.
!
of 49 64
Let us write the computation as one equation, and verify it. We have
Where
=Ionization energy of Hydrogen
=3.47E-39 (correct)
Equation 7.16
(0.000034574)(3.47E-39)=1.19972E-43
1.19972E-43^(1/3)
=4.932E-15
So as you can see equation 7.15 is correct. It says that carbon, the basis of life is in the ratio of
the solar magnetic field and the solar gravitational field.!
r =
(
q
2
p
μ
2
2GM
⊙
m
2
p
)
1/3
= 4.9324E − 15m
I
⊙
=
2B
⊙
R
⊙
μ
0
I
1AU
=
5.5362E12a mps
1.496E11m
= 37.0a m ps /m
1AU = r
e
μ =
Ioni zat ion En erg y
Magnet icFiel d St rength
=
2.18E − 18J
37.0A m ps /m
= 5.892E − 20
J ⋅ m
A
IE
H
I
⊙
1AU
=
2B
⊙
R
⊙
μ
0
1
r
e
μ
2
= (5.982E − 20)
2
= 3.47E − 39
μ
2
=
(IE
H
)
2
μ
2
0
r
2
e
4B
2
⊙
R
2
⊙
=
(2.18E − 18)
2
(12.56637E − 7)
2
(1.496E11)
2
4(5E − 3)
2
(6.957E8)
2
6r
p
=
(
q
2
p
2GM
⊙
m
2
p
⋅
(IE
H
)
2
μ
2
0
r
2
e
4B
2
⊙
R
2
⊙
)
1/3
6r
p
=
(1.6E − 19C )
2
2(6.67408E − 11N
m
2
kg
2
)(1.989E 30kg)(1.67E − 27)
2
⋅
(2.18E − 18J )
2
(12.56637E − 7T ⋅ m /A)
2
(1.496E11m)
2
4(5E − 3T )
2
(6.957E8m)
2
1/3
of 50 64
Thus we have a theoretical value for the radius of the proton:
Equation 7.17
And a theoretical value for its charge
Equation 7.18
Equation 7.19
And we have the radius of a proton in terms of the solar magnetic field at Earth
Equation 7.20
All of this based on the idea that the basis of their structure is in six-fold unfolding. Given our
constant k
Equation 7.21
And making the approximation we can with these equations eliminate and k
Equation 7.22
This has an accuracy of close to 88% because in equation 7.19 the charge of six protons is
predicted by the theory to be 5.72 protons, the rest of the equations are much more accurate, but
we seek to rectify that. We write this equation so we can have an equation that defines the solar
magnetic field by solving for .
To address the accuracy of the equation for the charge of a proton, equation 7.19, we ask what is
the culprit. We suggest it is . So we solve the equation for that to see by how much it is
off. It is:
Equation 7.23
We see it should be 7.79573E-11 and is actually 7.885E-11. The value we are using is 98.86785%
accurate, but we want to do better.
r
p
=
1
m
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
c
k
e
(
α
2
6
⋅
1
k
)
2
h 4π r
2
p
Gc
= q
2
q
q
p
= ⋅
9.1435E − 19C
1.602E − 19C
= 5.71proton s ≈ 6proton s
6r
p
=
(
q
2
p
2GM
⊙
m
2
p
⋅
(IE
H
)
2
μ
2
0
r
2
e
4B
2
⊙
R
2
⊙
)
1/3
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
1 + α = 1
q
p
31850496r
2
p
=
α
4
k
e
⋅
c
4
h
π
3
G
⋅
1
(N
A
𝔼)
2
h 4π
Gc
(IE
H
)
2
μ
2
0
r
2
e
GM
⊙
m
4
p
B
2
⊙
R
2
⊙
B
⊙
(α
4
)/36
α
4
36
= q
2
⋅
k
e
c
⋅ k
2
Gc
h 4π r
2
p
of 51 64
We now eliminate on the left in equation 7.22 with equation 7.17 to find the B field of the Sun
as described by the Earth orbit as the ground state:
Equation 7.24
Where N=31850496 is a perfect integer, that is has no values after the decimal. Equation 8 gives
a magnetic field strength of 4.73E-3 Teslas. Concerning equation 7.24
We want it to be more accurate. To do that we have to substitute for , the fraction
which is 39/5 and we have to formulate a theory for why this would be, though the discrepancy
could be in as its experimental value has the largest errors. Other possibilities are the
is right and that as well this is due to the radius of a proton having large errors, or even that it is
supposed to be the factor 8, which would be good to consider because 8-fold symmetry is very
dynamic, in particular in its role with beryllium 8 being a precursor to carbon in nuclear
synthesis by stars.
Because we are looking at 6 protons we had , and because of six-fold symmetry we had
and because the charge we had because of the 1/3 root on the right. So taking
and leaving the that came from substituting for k, and the as
well, and leaving the factors 2 and 4 because they describe the physical dynamics of the equation
we have:
Equation 7.25
Where
Or we can factor out all the numbers on the right the 2 and 4 in equation 7.25 and write
Equation 7.26
In which case N=31850496/4=7962624. We see the solar magnetic field is determined by the
radius of a hydrogen atom, its ionization energy, and the solar gravitational field, with the earth
orbit as the ground state.!
r
2
p
B
2
⊙
=
2
NR
2
H
⋅
(IE
H
)
2
μ
2
0
r
2
e
GM
⊙
m
2
p
R
2
⊙
⋅
α
4
c
3
k
e
h 4π
Gc
α
4
36
= q
2
⋅
k
e
c
⋅ k
2
Gc
h 4π r
2
p
(α
4
)/36
7
4
5
r
p
(α
4
)/36
6
2
α
2
/6
q = 6q
p
6
3
N = 6
7
= 279936
3
2
/4
2
8 = 2
3
B
2
⊙
=
2
NR
2
H
⋅
(IE
H
)
2
μ
2
0
r
2
e
2GM
⊙
m
2
p
4R
2
⊙
⋅
3
2
4
2
α
4
c
3
2
3
k
e
h 4π
Gc
N = 6
7
= 279936
B
2
⊙
=
1
NR
2
H
⋅
(IE
H
)
2
μ
2
0
r
2
e
GM
⊙
m
2
p
R
2
⊙
⋅
α
4
c
3
k
e
h π
Gc
r
e
of 52 64
Protons and neutrons pack in atoms in a way that can be seen from their electron clouds. Thus
since we can say the electron cloud of helium is spherical the nucleus packing is spherical. But
since the protons and and neutrons in the nucleus are moving (But only as far as their nuclear
wall) the packing is ordered but liquid.
I believe the reason for my finds of six-fold symmetry in both the radius of a proton and it its
charge can be explained by Buckminster Fuller’s vector equilibrium. It is the most
transformable straight-line geometry if you attach sticks with flexible corners, what Fuller calls a
flex-corner. See the following illustration.
Which means it zero-frequency for omni directional closest packing of spheres is 12 spheres,
which is carbon (six protons and six neutrons) which is the basis of life as we know it. Its
diameter is then six proton radii which we found provided for by the solar magnetic field. The
frequency is the number on any symmetrically concentric shell or layer and is given by
The more protons an element has the more neutrons it needs so the strong nuclear force in the
nucleus can overcome the mutual repulsion of protons, thus holding the atoms together.
10F
2
+ 2 = 10(1)
2
+ 2 = 12 = carbon
of 53 64
Thus, to go over that of the solar magnetic field again, proposed by Hannes Alfven (1942), at the
time there was no known mechanism for it, but he suggested the relative velocity between a
neutral gas and a plasma has a critical velocity at which the gas starts to ionize and that the
atoms or molecules will not exceed this velocity until the gas becomes almost fully ionized. The
additional energy put into the system goes into ionizing the gas instead of the velocity of the
atoms, and is roughly independent of pressure and magnetic field. Critical ionization velocity
has been recognized in the laboratory for some time. It is given by equating the kinetic energy of
the atoms to the ionization potential:
Alfven found:
Gas cloud enters Solar System
A neutral atom falls into Sun due to gravity
Motion is random, collisions happen
Temperature rises
At a distance from the Sun gas will ionize
m=atomic weight
He found for a gas cloud with average voltage 12 volts. An average atomic weight of 7, the is at
Jupiter.
That is atoms fall in towards the Sun and ionize at which point the solar magnetic field pushes
them out to Jupiter orbit where a halo forms from which planets can form.
1
2
mv
2
= eV
ion
r
i
= G
Mm
eV
ion
r
i
= G
Mm
eV
ion
= 13.5E10
m
V
ion
cm
V
ion
= volts
r
i
of 54 64
Thus if we are to suggest, as we have, that solar magnetic field provides for life, whose skeletons
are the hydrocarbons, that this has something to due with the solar wind which is plasma and
we can consider it a plasma that can move in neutral gas thus being subject to Alfven’s critical
velocity. The plasma is mostly electrons, protons, alpha particles kinetic energies between 0.5
and 10 keV. There are trace amounts of heavy ions of C, N, O, Ne Mg, Is, S, and Fe. We note the
C, N, O are the most abundant elements in life chemistry. The solar wind can reach velocities of
25o000-750000 m/s. Remember the critical velocity for hydrogen is 50,900 m/s.
of 55 64
Appendix 1 Kinetic Energies Moon and Earth
We suggested the second was a natural unit, and that, if it was, should be in the orbital
mechanics of the earth because the second comes from the calendar, which is based on the
orbital period of the year (1 year, 365.25 days) and the orbital period of the moon, and the
rotation of the Earth. We found that it was as the following:
Let’s show that…
To find the translational kinetic energy of the moon:
Distance from earth: 3.85E8m
Orbital period:
T=27.32 days=2.36E6seconds
v=1.025E3m/s~1000m/s
Mass: 7.34767E22kg
Use
E=3.67E28 Joules
To find the translational kinetic energy of the earth:
Distance from Sun: 1AU=1.496E11m
Orbital period: 1 year=3.1558E7 seconds
v=2.9785E4m/s, Earth mass: 5.9722E24kg
E=2.649E33 Joules
Earth day=(24)(60)(60)=86,400 seconds.
Using the Moon’s orbital velocity at aphelion, and Earth’s orbital velocity at perihelion we have:
Using the Moon’s orbital velocity at aphelion, and Earth’s orbital velocity at perihelion.
K E
moon
K E
earth
(Ear th Day) ≈ 1secon d
K . E . Moon = 3.67E 28J
K . E . Ear th = 2.649E 33j
2π (3.85E 8m) = 2.419E 9m
E =
1
2
mv
2
2π (1.496E11m) = 9.399E11m
K E
moon
=
1
2
(7.347673E 22kg)(966m /s)
2
= 3.428E 28J
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
of 56 64
Appendix 2 Van Der Waals Radius
Johannes Diderik van der Waals (1873) described more than just Ideal gases, which are gases
that behave according to kinetic-molecular theory, he described real gases which don’t. His
equation then, The Van der Waals equation, is a modification of the Ideal Gas Law which is:
Which is quite obvious. If you increase the temperature T, then the volume of the gas is going to
increase, and if it doesn’t then the pressure will, which is inversely proportional to volume.
However for a Real Gas, he assumed the particles are hard spheres, cannot be compressed
beyond a limit, and at close proximity to one another they interact and have a volume around
them that excludes one another, that is they have walls. He said
That is the volume of the real gas ( ) is equal to the volume of the ideal gas ( ) minus a
correction factor b. The volume of the particles is the number of particles ( ) times the volume
of one particle:
Thus there exists a sphere of radius 2r formed by two particles in contact where no other
particles can enter. It gives the correction factor
PV = n RT
V
R
= V
I
− b
V
R
V
I
n
n
4
3
π r
3
of 57 64
And the volume correction for n particles is
This is the volume correction to the Ideal Gas Law. The pressure correction says real gases
exhibit less pressure because their particles interact which is a net pulling by the bulk of
particles away from the container walls.
The reduction in pressure is proportional to by a factor a. We have for reduction of pressure
that
We substitute this into the Ideal Gas Law:
This can be written as a cubic
Which allows one to compute the critical conditions of liquefaction and to derive an expression
of the principle corresponding states. In the cubic form we have as the solution three volumes
which can be used for computing the volume at and below critical temperatures.
Thus the Van Der Waals radius is estimated
For hydrogen experimentally. Therefore with
We have described the derivation of radius of a hydrogen atom from the Van Der Waals
equations that we use to get
b = (4)
4
3
π r
3
nb = 4n ×
4
3
π r
3
n
2
v
2
P
I
= P
R
+ a
n
2
V
2
(
P + a
n
2
V
2
)
(
V − nb
)
= n RT
V
3
−
(
b +
RT
P
)
V
2
+
a
p
V −
a b
P
= 0
(
π +
3
φ
2
)
(3φ − 1) = 8τ : π =
P
P
c
, φ =
V
V
c
, τ =
T
T
c
4
3
π r
3
w
=
b
N
A
r
3
w
=
3
4π
⋅
b
N
A
b = 26.61
cm
3
m ol
N
A
= 6.02E 23
r
w
= 1.0967E − 8cm = 1.0967E − 10m
of 58 64
Where the Van Der Waals equations are
2
6
R
H
h
π α
2
m
2
p
GN
A
= 1secon d
(
P + a
n
2
V
2
)
(
V − nb
)
= n RT
V
3
−
(
b +
RT
P
)
V
2
+
a
p
V −
a b
P
= 0
4
3
π r
3
w
=
b
N
A
of 59 64
Appendix 3 Chandrasekhar Limit
The pressure of the outer shell of star balances with the outward pressure in the core of
the star (thermal pressure). Pressure is force per unit surface area thus…
is the mass of the core pulling in the mass of the shell and is the radius of the
core. The surface area of the star is that of a sphere, . We have
The thermal pressure countering the gravity is given by the ideal gas law PV=nRT (pressure
times volume of a gas such as hydrogen , which is all protons , is proportional to temperature.
The number of protons in the core is . We have
Where is the Boltzmann constant ( ). Since we must have
if the star is not to implode or explode
And we have the estimate for the temperature of the core of a star.
Fusion would not occur at the low temperature of a star like the Sun in that there would not be
enough energy for collisions, unless the potential Coulomb barrier can be overcome by quantum
mechanical tunneling. The collisions are given by the kinetic energy of the particles
. We have
The velocity v yields the minimum distance between protons as the De Broglie wavelength
P
gravit y
P =
F
A
F = m a = PA
P =
m a
A
m a = G
M
shell
M
core
r
2
core
M
core
M
shell
r
core
A = 4π r
2
core
P
gravit y
= G
M
shell
M
core
4π r
4
core
m
p
N
p
M
core
m
p
P
thermal
=
M
core
m
p
⋅
1
4
3
π r
3
core
k
B
T
core
k
B
1.380649E − 23J ⋅ K
−1
P
gravit y
= P
thermal
k
B
T
core
=
1
3
GM
shell
m
p
r
core
1
2
m
2
p
v
e
2
4πϵ
0
r
min
=
1
2
m
p
v
2
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Since the velocity is the root mean square velocity of the protons…
We have the temperature of the star is
This is another estimate. Since the mass of a star is its volume times its density
But for a star density varies with radius
If we take the derivative of both sides of the equation we have one of the equations of stellar
structure:
1.
The so-called conservation of mass equation. The force on the shell of the star is given by the
mass of the shell
Again for there to be balance gravitation pressure equals thermal pressure:
2.
Another equation of the equations of stellar structure. The so-called equation of hydrostatic
equilibrium. This can be written
If the star is an ideal gas the density of the star varies as where for a
monatomic gas and then
λ =
h
m
2
p
v
v
rms
=
3k
B
T
mp
T
min
=
(
e
2
4πϵ
0
)
2
m
p
3π
2
h
2
k
B
m =
4
3
π r
2
ρ
4π
∫
r
0
r
2
ρ(r)dr
dm(r)
dr
= 4π r
2
ρ(r)
F
g
= − G
m(r)4π r
2
ρ(r)
r
2
dr
dρ(r)
dr
= − G
m(r)ρ(r)
r
2
d
dr
(
r
2
ρ(r)
dP(r)
dr
)
= − 4π Gr
2
ρr
PV
γ
= con sta nt
γ = 5/3
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In stellar dynamics we write
So that
The abundance of hydrogen and helium in the universe are approximately 75% and 24%,
respectively. Thus for every 4He2+ there are 12H+ and 2+12 free electrons. We have
Ionized hydrogen and helium have and for the Sun because of high metal
content. Finally stars can be approximated as blackbody radiators (purely radiate) and as such
pressure is given in terms of temperature (Temperature is proportional to radiation energy):
There are three kinds of pressures that can be generated by a star: gas pressure, radiation
pressure, or degeneracy pressure.
A type of star that is stable, that is prevented from collapse by degeneracy pressure, is a so-called
white dwarf star. They are the remnant of giant stars that have depleted the their fusion fuel and
thereby collapsed under gravity but are kept from collapsing into black holes by thermal
pressure due to motion of the particles alone. Interestingly, they still shine almost as bright as a
star on the main sequence even though they are not doing fusion. It was the Indian physicist
Chandrasekhar who found the limit in mass for which a white dwarf will not have its gravity
overcome the degeneracy pressure and collapse. The non-relativistic equation is:
There are many resources available that derive this and you can find it in any textbook on
astrophysics in the chapters dealing with stellar physics, and I will leave the treatment of the
derivation to those works.
P ∝
1
V
γ
∝ ρ
5/3
N
V
=
ρ
μm
p
P
gas
=
ρ
μm
p
k
B
T
4 + 12
1 + 12 + 14
= 0.59
μ = 0.59
μ = 0.62
P
rad
=
4
3
⋅
σ
c
T
4
M ≤ 0.77
c
3
h
3
G
3
N
m
4
p
= 1.41 ⊙
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Appendix 4 Cosmology Overview
If the universe was born from a big explosion called “the big bang” then it is expanding and the
distances between galaxies is growing. This does not mean there is an edge to the universe —
space could continue on infinitely in all directions but there could be a horizon beyond which
there is no content.
There can be an observable edge to the universe, but it is more of a horizon, just as there is a
horizon to the ocean, we know beyond the horizon there is more water. Since the universe was
born 13.82 billion years ago in the Big Bang we can only see those galaxies whose light has taken
less than 13.82 billion years to reach us. These galaxies form a sphere around the Earth called
“the observable universe”. It is 92 billion light years in diameter because in its beginning the
universe inflated much faster than the speed of light.
Let’s discuss the types of possible universes.
The pivotal parameter describing the universe is 𝛀 (omega) which is the average matter density
of the universe divided by a critical value of that density. Whether omega is less than1, 1, or
greater than 1 determines whether the Universe is open, flat, or closed. If matter is mostly inert
as in the dust models, there is a particular fate for each omega. Since 1998 the observations in
supernovas indicate the universe is accelerating in its expansion. To explain this cosmologists
hypothesize the existence of dark energy which can be any field with negative entropy causing
negative pressure. If 𝛀<1 then the Universe is open and the fate of the universe is in its heat
death, the burning out of the last stars born. This is also true of a flat universe if it has the
hypothesized dark energy, because then it accelerates to escape collapse. The fate of the universe
depends on its density and today most evidence points to it will expand indefinitely. In this
scenario there is enough a supply of gas for stars to form for 10¹² to 10¹⁴ years. 1 to 100 trillion
years. That is the last stars born would die in 10¹⁴ years. There are models that suggest the
universe is eternal where as stars die more matter comes into existence from which more stars
are continuously forming.
Einstein’s equations give the relationship between spacetime and matter content of the
Universe. From this Friedmann gave us his equations that are at the heart of cosmology, which
are functions of time, the scale-factor R, the matter density of the universe , and the pressure p
due to radiation produced by the stars, or by the galaxies because they contain the stars:
Where
for a 3-sphere, for 3-space, and for a 3-pseudosphere. These Friedmann
equations give the critical density of the universe as
ρ
2
R′ ′
R
+
(
R′
R
)
2
+
κ
R
2
+ 8π p = 0
(
R′
R
)
2
+
κ
R
2
=
8πρ
3
R′ ≡
dR
dt
κ = + 1
κ = 0
κ = − 1
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Where is the Hubble constant which is the expansion rate of the universe. Current estimates
are that it is .
ρ
c
=
3H
2
0
8π G
H
0
71k m /s /Mpc
H
0
= 71,000m /s /Mpc
(3.26E6ly /Mpc)(9.461E15m /ly) = 3.1E 22m /Mpc
H
0
= (71000m /s)/(3.1E 22m) = 2.3E − 18s
−1
ρ
c
=
3(2.3E − 18)
2
8(3.141)(6.6741E − 11)
= 9.461E − 27kg /m
3
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The Author
